Ordinary Differential Equation - non-homogeneous

Question

The equation I'm given is

$$\frac{d^2y}{d x^2} + \frac{4}{x}\frac{dy}{dx} = x$$

I'm given that $u(x)=y'(x)$ and have to convert it into first order ODE for $u$ which (with the help of fellow members) I got the answer to be $$u=\frac{1}{6}x^2+\frac{C}{x^{4}}$$ which I believe is correct.

Now it asks me to find the general solution of the equation obtained for $u$ and then use that to find the general solution of the first (original) equation. I'm stuggling to find on what to do next, do I need to find a particular integral and if so, which equation do I use?

Answer 1

Note that $$y'=u \implies y(x)=\int u(x) dx$$

here is another solution

$$y''+\frac 4xy'=x$$ multiply both side by $x^4$ $$x^4y''+4x^3y'=x^5$$ $$(x^4y')=\int x^5 dx$$ $$x^4y'=\frac 16 x^6+K_1$$ $$y'=\frac {K_1}{x^4}+\frac 16x^2$$ Integrate $$y=\int \frac {K_1}{x^4}+\frac 16x^2dx$$ $$y(x)= \frac {C_1}{x^3}+\frac {x^3} {18}+C_2$$ With the initial conditions $$(y(1),y'(1))=(1/18,1/2) \implies (C_1,C_2)=(-1/9,1/9)$$ $$\boxed{y(x)=-\frac 1{9x^3}+\frac {x^3} {18}+\frac 19}$$

Answer 2

In this example, the differential equation takes the form $$\frac{1}{x^{4}} \, \frac{d}{dx} \left(x^{4} \, \frac{dy}{dx} \right) = x.$$ Now, \begin{align} \frac{1}{x^{4}} \, \frac{d}{dx} \left(x^{4} \, \frac{dy}{dx} \right) &= x \\ \frac{d}{dx} \left(x^{4} \, \frac{dy}{dx} \right) &= x^5 \\ x^{4} \, y' &= \frac{x^{6}}{6} + c_{1} \\ \frac{dy}{dx} &= \frac{x^{2}}{6} + \frac{c_{1}}{x^{4}} \\ y(x) &= \frac{x^{3}}{18} - \frac{c_{1}}{3 \, x^{3}} + c_{0}. \end{align}